In triangles \(
ABC\) and \(
PQR\), medians
\(
AD\) and
\(
PM\) correspond under the similarity \(\Delta ABC\) is similar to \(\Delta PQR\).. Prove that the medians are in the same ratio as the corresponding sides, i.e., \(\frac{AB}{PQ}=\frac{AD}{PM}\)
.
Proof:
Since \(AD\) is the median of \(\Delta ABC\)
\(BD =\) \( = \frac{1}{2}\)
Similarly, \(PM\) is the median \(\Delta PQR\)
\(QM = \) \(= \frac{1}{2}\)
Given \(\Delta ABC \sim \Delta PQR\).
Corresponding sides of similar triangle are proportional.
So, \(\frac{AB}{PQ} = \frac{BC}{QR}\)
\(\frac{AB}{PQ} = \frac{2BD}{2QM}\) (since \(AD\) and \(PM\) are medians)
\(\frac{AB}{PQ} = \frac{BD}{QM}\) - - - - - (1)
Since \(\Delta ABC \sim \Delta PQR\)
Corresponding angles of similar triangles are equal.
\(\angle B = \angle\) - - - - - (2)
Now, in \(\Delta ABD\) and \(\Delta PQM\)
\(\angle B = \angle Q\) (from (2))
\(\frac{AB}{PQ} = \frac{BD}{QM}\) (from (1))
Thus, \(\Delta ABD \sim \Delta PQM\) (by )
Since corresponding sides of similar triangles are proportional.
\(\frac{AB}{PQ} = \frac{AD}{PM}\)
Hence proved.